"for..of" javascript return all value but return a extra index of undefined

Question

Here is my code

function consec(string) {
  for (let letter of string) {
    console.log(letter);
  }
}

console.log(consec("zoo"));

the result is

"z"
"o"
"o"
undefined

This is the link of my code. Please feel free to correct. (It's my first to ask question through SO, I appreciate any advices :))

Last undefined is the return value of consec() function. Since you are not returing anything from consec(), it is displaying undefined. — Hassan Imam
– Hassan Imam, Commented Feb 9, 2018 at 7:31
undefinedhere is the return value from the function. The characters are logged in the function, and "undefined" is logged from the .log which calls consec because the function doesn't return anything. — Teemu
– Teemu, Commented Feb 9, 2018 at 7:31

Ankit Agarwal · Accepted Answer · 2018-02-09 07:33:43Z

6

Remove the console.log() from console.log(consec("zoo")); then you will not get that undefined printed in console as the last undefined is from the return value of consec("zoo"); which is undefined.

function consec(string) {
  for (let letter of string) {
    console.log(letter);
  }
}

consec("zoo");

Let's say if you have return value in function then you get that in console like this:

function consec(string) {
  for (let letter of string) {
    console.log(letter);
  }
  return("Finished");
}

console.log(consec("zoo"));

answered Feb 9, 2018 at 7:33

Ankit Agarwal

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